The discussion centers on a linear transformation T from R3 to R3, specifically defined by T(1, 0, 0) = (1, 0, -1), T(0, 1, 0) = (1, 0, -1), and T(0, 0, 1) = (1, 2, 2). The transformation can be represented using the matrix $$ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 2 \\ -1 & -1 & 2\\ \end{bmatrix} $$ which maps vectors in R3 to the subspace spanned by (1, 0, -1) and (1, 2, 2). The discussion also explores alternative matrix representations and confirms that switching basis vectors can yield valid transformations.
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A linear transformation can be fully described by its action on any basis. Can you see why?jolly_math said:Homework Statement:: Describe explicitly a linear transformation from R3 into R3 which has as its
range the subspace spanned by (1, 0, -1) and (1, 2, 2).
Relevant Equations:: linear transformation
"There is a linear transformation T from R3 to R3 such that T (1, 0, 0) = (1,0,−1), T(0,1,0) = (1,0,−1) and T(0,0,1) = (1,2,2)" - why is this the case?
Thank you.
Hall said:$$
\begin{bmatrix}
1 & 1 & 1 \\
0 & 0 & 2 \\
-1 & -1 & 2\\
\end{bmatrix}
\times
\begin{bmatrix}
x \\
y \\
z\\
\end{bmatrix}
$$
Yes, I think. I think even double columns of ##(1,2,2)## will also satisfy the given things.jolly_math said:For R3, would
$$
\begin{bmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
-1 & 2 & -1\\
\end{bmatrix}
\times
\begin{bmatrix}
x \\
y \\
z\\
\end{bmatrix}
$$
also work (switching which vector corresponds to each basis)? Thanks.